Question 1183353
</pre><pre>
Instead of doing yours for you, I'll do one step-by-step exactly like yours.
You can use it as a model to solve yours by.</pre>
A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point.
Degree 3	
Zeros  -4,4+(4square root5)i    
Solution Point
f(−2)  = −696

(a) Write the function in completely factored form.
f(x) = 
(b) Write the function in polynomial form.
f(x) =<pre> 
Solution step-by-step:
{{{"f(x)"=k(x-(-4)^"")(x-(4+4sqrt(5)*i)^"")(x-(4-4sqrt(5)*i)^"")}}}

{{{"f(x)"=k(x+4^"")(x-4-4sqrt(5)*i^"")(x-4+4sqrt(5)*i^"")}}}

[That will be the completely factored form, once we find k and substitute

{{{"f(x)"=k(x+4^"")((x-4)-4sqrt(5)*i^"")((x-4)+4sqrt(5)*i^"")}}}

The last two factors are conjugates, so we form their product which
is the difference of the squares of the terms:

{{{"f(x)"=k(x+4^"")((x-4)^2-(4sqrt(5)*i)^2)}}}

{{{"f(x)"=k(x+4^"")((x^2-8x+16)-16*5*i^2)}}}

{{{"f(x)"=k(x+4^"")(x^2-8x+16-80*(-1))}}}

{{{"f(x)"=k(x+4^"")(x^2-8x+16+80)}}}

{{{"f(x)"=k(x+4^"")(x^2-8x+96)}}}

{{{"f(x)"=k(x^3-8x^2+4x^2-32x+96x+384)}}}

{{{"f(x)"=k(x^3-4x^2+64x+384)}}}

{{{"f(x)"=k(x^3-4x^2+64x+384)}}}

{{{"f(-2)"=k((-2)^3-4(-2)^2+64(-2)+384)}}}

{{{-696=k((-8)-4(4)-128+384)}}}

{{{-696=k(-8-16-128+384)}}}

{{{-696=k(232)}}}

{{{(-696)/(232)=k}}}

{{{-3=k}}}

{{{"f(x)"=k(x^3-4x^2+64x+384)}}}

{{{"f(x)"=-3(x^3-4x^2+64x+384)}}}

{{{"f(x)"=-3x^3+12x^2-192x-1152)}}}

Completely factored form:

{{{"f(x)"=k(x+4^"")(x-4-4sqrt(5)*i^"")(x-4+4sqrt(5)*i^"")}}}

{{{"f(x)"=-3(x+4^"")(x-4-4sqrt(5)*i^"")(x-4+4sqrt(5)*i^"")}}}

Edwin</pre>